just a little googling would have solved your problem, for example read this about logical operators, like this? ITEproduction_2014.2015<-subset(ITEproduction_2014.2015,Date.Difference>3 & Date.Difference<40) ...

Create a column "F" with the formula =COUNTIF(B:B,B1) in the first row. Double-click the fill-down button (the bottom-right corner of the cell) to copy the formula for all your data. Then, click within the table and do CTRL+A followed by CTRL+L - when it asks if your data has headers,...

You first have to convert the column with the date values to datetime's with to_datetime(): df['datestart'] = pd.to_datetime(df['datestart'], coerce=True) This should normally parse the different formats flexibly (the coerce=True is important here to convert invalid dates to NaT). If you then want the year part of the dates, you can...

Your question is not very clear because the rules around what is "correct" are not defined. This returns the output you say you are expecting. declare @t table (firstname varchar(20),lastname varchar(20)) insert into @t values ('Mike','Odempsey') ,('Mike','O dempsey') ,('Mike','O''dempsey') ,('Mike','O"dempsey') ,('Steve','Royal') ,('Steve','Royal') ,('John','Royal Star') ,('John','Royal Star'); with CTE as( SELECT...

There are better ways to do this in R (Google the sample function), but to get every 10th row, assuming your data is called data: data[(1:100000)*10,] It is as easy as that. 1:10000 generates a list of numbers from 1 to 100000. Then those number are multiplied by 10, giving...

Your plan is fine. There is no risk in switching between recovery modes in your scenario. Your DB will be in Full mode after last backup. Only one note, if you need only 10% of data, wouldn't be faster to create new database (properly sized), copy over valid data, rename...

You are passing in kk to your strptime(...) which is a series where it is expecting a string. You can iterate over the rows in the series if you wish to print out each of the dates like so: for i in range(len(kk)): df_dt=datetime.datetime.strptime(kk[i],"%y%m%d%H") print(df_dt) which will give you 2014-10-21...